0200 - 0299
0290 - Word Pattern (Easy)
Problem Link
https://leetcode.com/problems/word-pattern/
Problem Statement
Given a pattern and a string s, find if s follows the same pattern.
Here follow means a full match, such that there is a bijection between a letter in pattern and a non-empty word in s.
Example 1:
Input: pattern = "abba", s = "dog cat cat dog"
Output: trueExample 2:
Input: pattern = "abba", s = "dog cat cat fish"
Output: falseExample 3:
Input: pattern = "aaaa", s = "dog cat cat dog"
Output: falseConstraints:
1 <= pattern.length <= 300patterncontains only lower-case English letters.1 <= s.length <= 3000scontains only lowercase English letters and spaces' '.sdoes not contain any leading or trailing spaces.- All the words in
sare separated by a single space.
Approach 1: Hashmap
class Solution {
public:
bool wordPattern(string pattern, string s) {
// convert s to a vector of strings
// e.g. "dog cat cat dog" -> ["dog", "cat", "cat", "dog"]
stringstream ss(s);
string word;
vector<string> words;
while (ss >> word) {
words.push_back(word);
}
// the size of words needs to be same as that of pattern
// e.g. words = ["xxx"], pattern = "xxx"
if (words.size() != pattern.size()) {
return false;
}
// for each word in words ...
// char in pattern -> word
// e.g. a -> dog
// e.g. b -> cat
unordered_map<char, string> m;
set<string> used;
for (int i = 0; i < words.size(); i++) {
// check if map the pattern
if (m.count(pattern[i])) {
// if pattern[i] exists in the hashmap,
// then we need to make sure that the word is correct
if (m[pattern[i]] != words[i]) {
return false;
}
} else {
// each word can only map to one pattern
// e.g. pattern = "ab", s = "dog dog"
if (used.find(words[i]) != used.end()) {
return false;
}
// if not, then map it
// e.g. a -> dog
m[pattern[i]] = words[i];
used.insert(words[i]);
}
}
return true;
}
};